# Estimating the concentration of nanoparticles from the particle size data

In this section of the module students calculate the nanoparticle concentration for each synthetic condition. First, they calculate the average number of gold atoms per nanoparticle (N), then the total number of particles based on the moles of gold used in the reaction and from there, the actual nanoparticle concentration expressed in molarity.

Q23. According to data from Table 2, nanoparticles synthesized with a 2:1 citrate to tetrachloroauric acid ratio and pH 5.4 have a Feret’s diameter of 21.7 nm. What is the value of N?

The relationship between the average number of gold atoms (N) per nanoparticle and the particle diameter (D) is provided by equation $$\ref{eq. 1}$$:

$\mathrm{N = \dfrac{π \left(19.3 \dfrac{g}{cm^3}\right) \mathit{D}^3}{6\left(197 \dfrac{g}{mol}\right)}} \label{eq. 1}$

This equation assumes a spherical shape and a uniform face-centered cubic (fcc) structure. In equation $$\ref{eq. 1}$$, 19.3 g/cm3 is the density for fcc gold and 197 g/mol is the gold atomic mass. For D = 21.7 nm,

$\mathrm{N = \dfrac{π (19.3\: g/cm^3)\: D^3}{6 (197\: g/mol)}= 316,946\: atoms/nanoparticle}\nonumber$

Q24. What is the total number of gold atoms (NTotal) in 50.0 mL of a 0.25 mM solution of HAuCl4?

The total number of gold atoms in 50.0 mL of a 0.25nM solution of HAuCl4 can be estimated by calculating the number of moles of gold and multiplying by Avogadro’s number:

$\mathrm{0.050\: L \times \left (0.00025\: \frac{mol}{L} \right ) \times \left (\dfrac{1\: mol\: Au}{1\: mol\: HAuCl_4} \right ) \times \left (6.02 \times 10^{23}\: \dfrac{atoms}{mole} \right ) = 7.5 \times 10^{19}\: atoms}\nonumber$

Q25. What is the molar concentration of nanoparticles in this solution?

The molar concentration of a nanoparticle solution can be estimated by dividing the total number of gold atoms (Ntotal) equivalent to the amount of auric acid added to the reaction volume by the average number of gold atoms (N).

$\mathrm{C =\dfrac{N_{total}}{NVN_A}}\nonumber$

This is equivalent to the equation

$\mathrm{C = \dfrac{moles\: from\: reaction}{N \times Volume}}\nonumber$

$\mathrm{C = \dfrac{1.25 \times 10^{-5}\: mol\: Au}{316,946\: atoms/NP \times 0.05\:L} =7.9 \times 10^{-10}\: mol/L}\nonumber$

Following the example provided above, students should repeat the calculations for all particles and complete all data in Table 4.